Problem 13 - Calculus Single Variable

Question

A function \(f\) is defined piecewise on an interval \(I=[a, b] .\) Find the area of the region that is between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. $$ f(x)=\left\\{\begin{array}{cl} -x^{2} & \text { if }-3 \leq x \leq 1 \\ 2 x-3 & \text { if } 1

Step-by-Step Solution

Answer

The total area is \( \frac{44}{3} \).

Step 1: Identify the Intervals and Functions

The function is piecewise defined for two intervals: 1. For \( -3 \leq x \leq 1 \): \( f(x) = -x^2 \) 2. For \( 1 < x \leq 4 \): \( f(x) = 2x - 3 \). The overall interval is \( I = [-3, 4] \). To find the area under the graph and above the x-axis, we need to integrate \( f(x) \) over these intervals.

Step 2: Set Up the Integrals for Each Interval

For the first interval \([-3, 1]\), the integral is \(\int_{-3}^{1} -x^2 \, dx \). For the second interval \((1, 4]\), set up the integral as: \( \int_{1}^{4} (2x - 3) \, dx \). These integrals will give the areas under \( f(x) \) in their respective intervals.

Step 3: Calculate the First Integral

Calculate the integral for the first interval: \( \int_{-3}^{1} -x^2 \, dx = \left[ \frac{-x^3}{3} \right]_{-3}^{1} = \left( \frac{-1^3}{3} \right) - \left( \frac{-(-3)^3}{3} \right) = -\frac{1}{3} + 9 = \frac{26}{3}. \)

Step 4: Calculate the Second Integral

Calculate the integral for the second interval: \( \int_{1}^{4} (2x - 3) \, dx = \left[ x^2 - 3x \right]_{1}^{4} = ((4^2 - 3 imes 4) - (1^2 - 3 imes 1)) = (16 - 12) - (1 - 3) = 4 + 2 = 6. \)

Step 5: Sum the Areas for Total Area

The total area between the function and the x-axis from \( x = -3 \) to \( x = 4 \) is the sum of both calculated areas: \( \frac{26}{3} + 6 = \frac{26}{3} + \frac{18}{3} = \frac{44}{3}. \)

Key Concepts

Definite Integrals

In calculus, definite integrals are used to calculate the exact area under the curve of a function within a specified interval.
Unlike indefinite integrals, which produce a general form of an antiderivative, definite integrals provide a specific numerical value, representing this area.
The process involves integrating the function over a specific interval \([a, b]\). This means computing the integral between these two points to find the bounded area. This is often visualized as the area between the curve and the x-axis, from the start of the interval to the end.

For example, when calculating the area for a piecewise function, the function is split into manageable sections, each defined by its interval. Each interval is then integrated separately using definite integrals.

• Calculate each definite integral within its respective interval.
• Add up the integrals of individual sections to obtain the overall area under the curve.
  

This technique can be especially useful when dealing with complex functions encompassing multiple expressions or segments.

Piecewise Functions

Piecewise functions are functions defined by multiple expressions, each corresponding to different parts of the function's domain.
This means that depending on the input value, a different expression may be applied to obtain the output.
In the given exercise, the function has two distinct expressions: \(-x^2\) when \(-3 \leq x \leq 1\), and \(2x - 3\) when \(1 < x \leq 4\).

A few key points about piecewise functions include:

• The need to carefully identify the intervals each expression is valid for, as this influences the integration process.
  
• Ensuring continuity, meaning each piece smoothly connects with the next, especially at the points where they meet (though not always required, it's often desirable).
  
• Breaking down the problem into smaller, manageable pieces simplifies the integration of complicated functions.
  

To solve integrals or any related problems involving piecewise functions, you need to handle each part separately. This method aids in capturing the behavior and attributes of the function on different segments of the domain.

Area under a Curve

Finding the area under a curve is a practical application of integration that provides useful information in various fields like physics, engineering, and statistics.
This concept involves calculating the section of the graph of a function that lies between the curve itself and the x-axis.
The definite integral of the function over a given interval gives the precise measure of this area.

In the context of the exercise, the area under the piecewise function is calculated in two main parts:

• First, find the area under each separate expression over its defined interval.
  
• Then, sum these individual areas to achieve the total area between the curve and the x-axis from \(-3\) to \(4\).
  

It's important to apply the integral on each segment correctly and ensure the addition of areas reflects any crossings of the x-axis (if the curve dips below it).
This technique effectively helps in determining how much space the function takes up relative to the axis and can illustrate trends and behaviors of different physical phenomena.